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Tom [10]
3 years ago
11

Control variables are parts of the experiment that you try to:

Chemistry
2 answers:
Katena32 [7]3 years ago
6 0

Answer:

A control variable in scientific experimentation is an experimental element which is constant and unchanged throughout the course of the investigation.

Hope this helped :)

maks197457 [2]3 years ago
3 0

Answer:

study, you do not change them as they are meant to be controls not like independent variables.

Explanation:

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Once again rocks and/or minerals
sergeinik [125]

Answer:

you can classify a mineral by its appearance and other properties. The presence of a mineral is defined by the color and luster, and the color of the powdered mineral is described by the band. Every mineral has a distinctive density. To compare the hardness of rocks, the Mohs Hardness Scale is used.

Explanation:

5 0
3 years ago
Read 2 more answers
2. During asexual reproduction in paramecia, a single paramecium becomes two new
zloy xaker [14]

Answer:

ion no

Explanation:

ion no

4 0
3 years ago
What is the mass of a bar of aluminum with length 5.3cm, width 0.32 cm and height 2.34cm? The density of aluminum is 2.70g/cm3.
Tatiana [17]

Answer:

<h2>mass = 10.72 g</h2>

Explanation:

Density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

Making mass the subject we have

mass = Density \times volume

From the question

Density = 2.70 g/cm³

We assume that the aluminum is a cuboid

and volume of a cuboid is given by

Volume = length × width × height

length = 5.3 cm

width = 0.32 cm

height = 2.34 cm

Volume = 5.3 × 0.32 × 2.34 = 3.97 cm³

Substitute the values into the above and solve for the mass

We have

mass = 2.70 × 3.97

We have the final answer as

<h3>mass = 10.72 g</h3>

Hope this helps you

6 0
3 years ago
Water glass is found in
klemol [59]

Answer:

liquid form

Explanation:

am i right? if right like

5 0
3 years ago
Read 2 more answers
CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the
nirvana33 [79]

Answer:

The answer to your question is   P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L                        Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C               Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa                    Pressure 2 = P2 =  ?

Process

- To solve this problem use the Combined gas law.

                     P1V1/T1 = P2V2/T2

-Solve for P2

                     P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

                       1000 ml -------------------- 1 l

                         855 ml --------------------  x

                         x = (855 x 1) / 1000

                         x = 0.855 l

-Substitution

                    P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

                    P2 = 377600 / 141.075

-Result

                   P2 = 2676.6 kPa

3 0
3 years ago
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